Problem: Solve for $z$, $ \dfrac{3z - 2}{2z + 1} = \dfrac{1}{9} $
Answer: Multiply both sides of the equation by $2z + 1$ $ 3z - 2 = \dfrac{2z + 1}{9} $ Multiply both sides of the equation by $9$ $ 9(3z - 2) = 2z + 1 $ $27z - 18 = 2z + 1$ $25z - 18 = 1$ $25z = 19$ $z = \dfrac{19}{25}$